Solve the following quadratic function and leave the answer as an exact value. (3x – 1) 2 = 2

Question

1. Find the equation of the quadratic function that has roots -8 and 2 and goes through the point (-3, 11)

2 The height of a ball thrown from a balcony can be modeled by the function h(t)= -5t2 + 30t + 35, where h(t) is the height above the ground, in metres, at time t seconds after it is thrown.

a. Write h(t) in vertex form.

b) What is the maximum height of the ball?

c) When does the ball reach its max. height?

d) Domain & Range

3 Let ℎ(𝑥) = −𝑥 2 + 9𝑥 − 12 𝑎𝑛𝑑 𝑔(𝑥) = −7𝑥 2 + 8

a) Find x when 𝒉(−𝟑) = g(x)

b) When g(x) = 10x, solve for x by factoring

4 Solve the following quadratic function and leave the answer as an exact value. (3x – 1) 2 = 2

5 Find the equation of the quadratic function that has the roots 2 + √6 and 2- √6, and passes through the point (7, -3)

6 Determine the values of k for which the function f(x) = 𝒙 𝟐 − 𝟐𝒙 + 𝟒𝒌𝒙 + 𝟗 = 𝟎 has two real roots. Check this value in the original equation.

Answer 1

(3x-1)^2 = 2

3x-1 = + or - sqr2

3x = 1+or-sqr2

x = 1/3+or-sqr2/3

x = about .80473 or -0.1381

x=-8 and 2 through (-3,11)

a(x+8)(x-2) = a(x^2 +6x -16) = y

11= a(9-18-16) = -25a

a = -11/25

f(x) = (-11/25)(x^2 +6x -16)

f(x) = -11x^2/25 -66x/25 + 176/25

h(t) =-5t^2 + 30t + 35

-5t^2 +30t +35=0

t^2 -6t -7=0

t^2 -6t +9 = 7+9 =16

(t-3)^2 = 16

t = 3 + or - 4 = -1 or 7 = the t intercepts

vertex is (3,80)

h(t) = (t-3)^2 +80

maximum height =h(3) = 80 meters at 3 seconds

range is [0,80]

domain is [0, 7]

h(x) =-x^2+9x-12

g(x) = -7x^2 + 8

h(-3) = g(x)

-9-27-12 = -7x^2 + 8

7x^2 = 8+9+27+12=56

x^2 =56/7 = 8

x=+ or - 2sqr2

x=2+sqr6

x=2-sqr6

x-2=sqr6

x^2-4x+4 =6

a(x^2-4x-2) = 0

through (7,-3)

-3 =a(49-28-2)=19a

a=-3/19

f(x) =(-3/19)(x^2-4x-2)

f(x)= -3x^2/19 +12x/19 +6/19

x^2-2x+4kx+9=0

k when there are two real roots

when the discriminant is >0

b^2 -4ac>0

b=4k-2, a=1, c=9

(4k-2)^2-4(9)>0

16k^2 -16k+4 -36>0

4k^2-4k-8

k^2-k-2 >0

(k-2)(k+1)>0

either both factors are positive or both negative

k>2 and k>-1,so k has to be >2

or

k<2 and k<-1, so k has to be <-1

either k<-1 or k>2 to have 2 real roots

k can't be in the interval [-1,2] if k=-1 or 2, there's only one real root. if -1<k<2 there's two imaginary roots